Prove that
Unbaised estimator of $\ e^{- 2 \lambda } $ is t(x) = $ \ (-1)^x $ ,
when x follows poisson distribution with parameter $\lambda $
Sorry to ask the direct question but I need a starting hint to solve the question , actually last time I studied statistical inference was 1 year back , I would be great if anyone could just give some starting steps .
I'll provide a step by step answer. In case you want to try it yourself, please stop at each point and try from there.
An unbiased estimator, $t(x)$, of a quantity $\theta$ (which in this case is $\theta = e^{-2\lambda}$), is such that: $$ \mathbb{E}t(x) = \theta = e^{-2\lambda} $$
we know $t(x)=(-1)^{x}$ and $x\sim \text{Poisson}(\lambda)$. Then we want to compute the expected value of this quantity: $$ \mathbb{E}\left[(-1)^x\right]. $$ Recall that expected value (for a discrete random variable) is obtained by summing all possible values the random variable can take multiplied by their probability. Recall further that if $x\sim \text{Poisson}(\lambda)$ then $P(x=i)=\frac{\lambda^{i}e^{-\lambda}}{i!}$. Thus, just using definitions: $$ $$ $$ \mathbb{E}\left[(-1)^x\right] = \sum_{i=0}^{\infty} (-1)^{i}P(x=i) $$
Using the Poisson law, we have $$ \sum_{i=0}^{\infty} (-1)^{i}P(x=i) = \sum_{i=0}^{\infty} (-1)^{i}\frac{\lambda^{i}e^{-\lambda}}{i!}=e^{-\lambda}-\lambda e^{-\lambda}+\lambda^{2} \frac{e^{-\lambda}}{2!}-\lambda^{3} \frac{e^{-\lambda}}{3!}+\ldots$$
Note that we can further simplify by taking common factor $e^{-\lambda}$ $$ e^{-\lambda}\left(1-\lambda+ \frac{\lambda^{2}}{2!}- \frac{\lambda^{3}}{3!}+\ldots\right) $$ Recall Taylor series? You can use them to rewrite $e^{-\lambda}$ around $0$ (which is actually the definition of exponential): $$ e^{-\lambda}= e^{-0} +\frac{\partial e^{-\lambda}}{\partial\lambda}|_{\lambda=0}\frac{(\lambda-0)}{1!}+\frac{\partial^{2} e^{-\lambda}}{\partial\lambda^{2}}|_{\lambda=0}\frac{(\lambda-0)^{2}}{2!} \ldots$$ solving the derivatives, you see that the above is: $$ e^{-\lambda}= 1 -\lambda+\frac{\lambda^{2}}{2!}-\frac{\lambda^{3}}{3!} \ldots$$ So, we use this fact to rewrite: $$ e^{-\lambda}\left(1-\lambda+ \frac{\lambda^{2}}{2!}- \frac{\lambda^{3}}{3!}+\ldots\right)=e^{-\lambda}e^{-\lambda}=e^{-2\lambda}. $$
Remember where we started from? $$ \mathbb{E}t(x) = e^{-\lambda}\left(1-\lambda+ \frac{\lambda^{2}}{2!}- \frac{\lambda^{3}}{3!}+\ldots\right)=e^{-\lambda}e^{-\lambda}=e^{-2\lambda} $$ Done!
If you want to show it's unbiased then you need to show it has the correct expectation. Then your hint is that you're going to want to think about the probability generating function of a Poisson. Hopefully that's enough.
E $\ (-1)^{x} $ = $ \frac {\sum_{x=0}^{ \infty} (-1)^{x} \ e^{- \lambda } \lambda^{x} }{\ x! } $
= $ e^{- \lambda } $ $ \frac {\sum_{x=0}^{ \infty } (- \lambda )^{x}}{ x !} $ { this is the property of exponential function }
= $ e^{- \lambda } $ $ e^{- \lambda } $
= $ e^{- 2\lambda } $
